Hey! Combinatorics + Inequality!

Let there be $w$ white balls and $b$ blue balls. The probability that both balls are of the same color is $\frac{1}{2}$ , so the complement event that both balls have different colors is $1 - \frac{1}{2} = \frac{1}{2}$ . But the probability that both balls have different colors is simply $\dfrac{wb}{\binom{w+b}{2}}$ . Thus,

$\dfrac{wb}{\binom{w+b}{2}} = \dfrac{1}{2}$

$\dfrac{wb}{(w+b)^2 - (w+b)} = \dfrac{1}{4}$

$(w+b)^2 - (w+b) = 4wb$

$(w+b)^2 - 4wb = w+b$

$(w-b)^2 = w+b$

Clearly it's better to take $w-b \ge 0$ to maximize $w$ (taking $w-b \le 0$ maximizes $b$ instead), so $w-b = \sqrt{w+b}$ instead of $-\sqrt{w+b}$ .

We want to maximize $w = \frac{(w+b) + (w-b)}{2} = \frac{(w+b) + \sqrt{w+b}}{2}$ . Since the latter is a monotonically increasing function on $w+b$ , we need to maximize $w+b$ . But we are given that $w+b \le 2009$ and that it is the square of an integer (namely the square of $w-b$ ). Thus we obtain $w+b = 44^2 = 1936$ , as the next square is $45^2 = 2025 > 2009$ , and so $w = \frac{1936 + \sqrt{1936}}{2} = \boxed{990}$ .

Let there be $n$ balls, of which $w$ are white and $n-w$ are blue.

• The probability of picking two white balls is $P(2W) = \frac{w(w-1)}{n(n-1)}$
• The probability of picking two blue balls is $P(2B) = \frac{(n-w)(n-w-1)}{n(n-1)}$
• The probability of picking two balls that are the same color is $P = P(2W) + P(2B) = \frac{w(w-1) + (n-w)(n-w-1)}{n(n-1)} = \frac{1}{2}$

We would like to express $w$ as a function of $n$ . The above expression simplifies:

• $4w^{2} - 4w \cdot n + n\cdot(n-1) = 0$

We can find acceptable values of $w$ by using the quadratic formula:

• $w = \frac{4n \pm \sqrt{16n^{2} - 16n(n-1)}}{8}$

The above expression simplifies:

• $w = \frac{n\pm\sqrt{n}}{2}$

This tells us two things:

• The largest possible value for $w$ increases with increasing $n$
• $n$ must be a perfect square

Therefore, n must be the largest perfect square that is less than or equal to 2009. This is $44^{2} = 1936$ .

So:

• $n = 1936$
• $w = \frac{n\pm\sqrt{n}}{2}$ , so the largest possible value for $w$ is $\boxed{990}$

Just to check our work, we can calculate the probability to pick two white balls or two blue balls under this configuration:

• $P = P(2W) + P(2B) = \frac{w(w-1) + (n-w)(n-w-1)}{n(n-1)}$
• $P = \frac{990\cdot 989 + 946\cdot945}{1936\cdot 1935} = \frac{1}{2}$